Polar Zonohedra

Zones, Zone Axes and Zonohedra

A zone is a family of faces of a polyhedron that are all parallel to a
given direction in space, called the zone axis. For example, the faces of
a prism make up a zone whose zone axis is parallel to the axis of the prism.
Only the orientation of the zone axis is important; it has no specific location.
In the polyhedra below, the green and blue faces make up a zone with the zone
axis vertical.

In a trivial sense, every pair of faces on a polyhedron can be thought of as
a zone, with the zone axis parallel to their mutual edge. But usually the term
zone is applied only to groups of faces that make up a large part of a
polyhedron, especially if the faces are related by symmetry.

The solids above illustrate some other zone concepts. The four red faces on
the cube at left obviously all belong to a zone, with the zone axis vertical. So
do the four red octagon faces of the truncated cube to its right. If we continue
truncating the cube, we get the cuboctahedron (third from left) which is halfway
between a cube and an octahedron. The four red faces do not share edges, but
they still constitute a zone with a vertical zone axis. So do the four red
faces in the truncated octahedron (far right). So we can see that the faces in a
zone need not intersect at all, as long as they are all parallel to the same
line in space.

The truncated octahedron consists of squares and hexagons. The top green face
and the hexagon below it intersect in a horizontal line, but the top hexagon
intersects with the hexagon below it in a horizontal line also. That hexagon in
turn intersects the bottom face in a horizontal line, and so on up the back of
the polyhedron. We have a set of faces completely girdling the polyhedron, all
parallel to a horizontal line. They make up a zone. Faces in a zone need not be
identical and need not be limited to four sides.

We can see that if a face adjoins two neighboring faces of the same zone, it
will have a pair of parallel edges. Solids where every edge is parallel to a
zone axis are called zonohedra.

Polar Zonohedra

Polar zonohedra have a single principal symmetry axis. Below we see the 3- 4-
and 5- fold polar zonohedra.

Each zonohedron has n edges meeting at the top vertex. If these are the only
edges allowed, then the faces around the vertex must be rhombuses. They in turn
define the angles of the next face, which must also be a rhombus, and so on.

There is no rule regarding scale along the polar axis. The three-fold
zonohedron is a rhombohedron, of which a cube is a special case. One special
case of the four-fold zonohedron is the rhombic dodecahedron. But if the
polyhedron is stretched a bit so the equatorial faces are squares, the top and
bottom sets of faces are 60-120 rhombuses.

We could stretch the five-fold zonohedron a bit so that it is girdled by a
zig-zag belt of squares. If we truncate the top and bottom pyramids, we are
tempted to think we have a solid made of regular polygons. But alas, no - the
triangles are not quite equilateral. But a cardboard model looks quite
convincing and attractive.

We can see that each zonohedron has n faces around each vertex and n-1 bands
of faces circling it, so a polar zonohedron has n(n-1) faces.

Viewed along the polar axis (above), polar zonohedra form lovely rosettes.
Each ring of rhombuses in turn dicates the next. If n is even, the limit of the
polyhedron occurs when the angle between intersecting rhombus edges equals 180
degrees. The equatorial cross-section of the zonohedron has n sides. If n is
odd, the limit of the polyhedron occurs when the angle between intersecting
rhombus edges exceeds 180 degrees. The equatorial cross-section of the
zonohedron has 2n sides.

Here we can see some basic geometry of zonohedra. At top left, we can see
that the i-th ring of rhombuses has angle iA, where A = 360/n. This is true for
any n. We'd also like to find a formula for the radius of each ring of vertices.
Start numbering the rings with zero at the center. The formula is easiest to see
for even n (top right). We can see that OE' is the diameter of a 2n-gon, and O'E'
is the diameter of a 2n-gon twice as large. If we define R=O'E', then we have
O'A'=2Rsin(A/2), O'B'=2Rsin(2A/2), O'C'=2Rsin(3A/2), and so on. If we define R1
= OA, R2 = OB, and so on, it's obvious that R1=O'A'/2, and so on. Therefore, Ri
= Rsin(iA/2). When Ri = R, sin(iA/2)=1, iA/2=90=180i/n, and i=n/2.

It's a bit harder to see the geometry for n odd (bottom) because the edges in
the rosette don't form replicas of the equatorial section, but the relationship
is the same.

Now let's look in the axial direction. If we define edge length as 1, and let
the angle the edge makes with the zone axis = Q, then the projection along the
axis of the edge is h. If we consider the edge at the top vertex, then the
projection normal to the axis is R1.

Since all the edges are parallel to one of the edges at the top vertex:

All edges have the same length projected on the polar axis

All the vertices lie on equally spaced planes, and since

Ri = Rsin(iA/2)

The axial section of the zonohedron is a sine curve.

As a fringe benefit, the formulas for vertex radius work for all values of
i. When i = n, Ri=sin(nA/2)=sin(n 360/n /2)=sin(180) = 0, and we are at the
bottom vertex. Also, for n odd, R(n-1)/2=R(n+1)/2. For example, when n=5,
R2=R3.

The one final task is to get the face angles. We do this at right above. If
Vi equals the vertex angle of a face in the i-th ring, then we find by
straightforward trigonometry that SinVi/2=Rsin(A/2)sin(iA/2). At first glance,
this looks strange. Where did angle Q go? Well, it turns out that angle Q
determines R1, and hence R.

In Terms of Edge Length = 1

Length L of polar axis = nh = ncosQ

Maximum radius R = sinQ/sin(A/2)

SinVi/2 = sinQsin(iA/2)

In Terms of Maximum Radius R

Edge = Rsin(A/2)/sinQ

Length L of polar axis = nh = ncosQ

SinVi/2=Rsin(A/2)sin(iA/2)

In Terms of Maximum Radius R and Polar Axis L

CosQ = L/n; calculate SinQ

Edge = Rsin(A/2)/sinQ

SinVi/2=Rsin(A/2)sin(iA/2)

Above is the 10-zonohedron, scaled so its polar axis and
diameter are about equal. In this view the successive rings of rhombus
faces are colored.

Here the polyhedron is colored so that faces of a given zone
have the same color.

Some Special Cases

n=3

Edge=1, A=120, A/2=60, sin(A/2)=sqrt(3)/2

Sin(V1/2) = Rsin(A/2)sin(A/2) = 3R/4

Sin(V2/2) = Rsin(A/2)sin(2A/2), but sin120 = sin 60 so V2=V1

If V1=90 (a cube), Sin(V1/2)=sqrt(2)/2 and R=2sqrt(2)/3

Since R = sinQ/sin(A/2), sinQ=Rsin(A/2)=Rsqrt(3)/2=sqrt(6)/3

SinQ=sqrt(6)/3 implies CosQ=sqrt(3)/3

L = 3cosQ = sqrt(3) - the correct length for a cube.

n=4

A=90, A/2=45, Sin(A/2)=sqrt(2)/2

Case 1: V2=90.

SinV2/2=Rsin(A/2)sin(2A/2)=Rsqrt(2)/2

sqrt(2)/2=Rsqrt(2)/2, hence R=1

SinV1/2=Rsin(A/2)sin(A/2)=R/2=1/2

Thus V1/2=30 and V1=60

Thus this solid has square equatorial faces and 60-120 top and bottom
faces, as described earlier.

Case 2: isohedral faces. V1=V2 does not work, so solve for V1=180-V2

SinV1/2=Rsin(A/2)sin(A/2)=R/2

SinV2/2=Rsin(A/2)sin(2A/2)=Rsqrt(2)/2=sqrt(2)sin(V1/2)

Sin(V1/2)=sin((180-V2)/2)=sin(90-V2/2)=cos(V2/2)

SinV2/2=sqrt(2)sin(V1/2), so sin2(v2/2)=2sin2(v1/2)

cos2(v2/2)=1-2sin2(v1/2)=sin2(v1/2)

Thus sin2(v1/2)=1/3, and sin(v1/2)=sqrt(3)/3

V1/2=35.26, thus V1=70.53, the angles for a rhombic dodecahedron.

Extended Zonohedra

Let's try the method above for n=5, to see if there are any isohedral
zonohedra.

The diameter, 2R = 3.0434, but the length along the polar axis is only 2.236.
We expect a rhombic triacontahedron to be roughly equidimensional. Wassup?

The solution is correct. It turns out there are several polyhedra that can be
constructed with the same face as the rhombic triacontahedron. The solid we have
actually constructed looks like this:

Recall that we discovered that a polar zonohedron has n(n-1)
faces. For n=5, there are 20 faces. There is a rosette of five faces
around each vertex (red, yellow, orange and green) and a zigzag band
around the middle (purple)

We can extend a zonohedron by adding a zone. We can separate the
polyhedron along any set of edges, pull the two pieces apart, and join the two
halves with equal and parallel edges.

The rhombic traicontahedron is actually made by extending
the basic polyhedron above. We separate the zigzag band of purple faces,
pull the two halves apart parallel to the polar axis by one edge length,
and add a new set of edges. We add a girdle of ten faces around the
equator. Note that these are identical to all the other faces.

Note that this particular extension will only work if n is odd.

Diminished Zonohedra

We can also diminish a zonohedron by removing any
zone. We can get a third polyhedron with rhombic triacontahedron faces by
removing a zone from the 20-face zonohedron above. It looks like this and
has 12 faces, but different symmetry from the regular rhombic
dodecahedron.

These three shapes, the rhombic dodecahedron, and the general
rhombohedron are the only convex solids with congruent rhombic faces.